BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

xy plane triangle

Expert replies
by GmatKiss » Sat Aug 06, 2011 12:03 pm
A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

90

900

8100

6480

10000
Join the discussion
Source: — Problem Solving |

by maihuna » Sat Aug 06, 2011 12:41 pm
AB parallel to X-axis so AB can be any of -3->5 and 2-11 i.e. a total of : 9*10 = 90.
Taken A, B can be choosen in remaining 8 ways and C in remaining 9 ways(one already taken by A)

so a total of : 90*8*9= 6480
GmatKiss wrote:A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

90

900

8100

6480

10000
Charged up again to beat the beast :)
Join the discussion

by GMATGuruNY » Sat Aug 06, 2011 1:03 pm
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by suruchia » Sun Aug 07, 2011 8:52 am
1. AB has to be parallel to x-axis...the range for x-axis for this question is -3 ≤ x ≤ 5..i.e 9 points.......therefore,
ways to choose 2 points out of 9..and order matters = 9 P 2 = 9! / 7 ! = 72

And each of the above permutation can happen in 10 ways ( consider y axis also)...So total ways for side AB = 72*10 = 720


2. As the right angle has to be at point A, the point C has to be directly above or directly below the point A, as the Y range is 2 ≤ y ≤11 i.e. 10 points...and A has already taken 1 point....C can take the remaining 9 points.

So Total no. of triangles that can be formed = 720 * 9 = 6480
GmatKiss wrote:A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

90

900

8100

6480

10000
Join the discussion